Mathematics

[Anileve]

Using 1,3,4 and 6 and only the four basic operations get to 24. Use each number only once.

Correct me if I am wrong, but why in the world would you want to complicate a simple solution just multiply 6X4 and get it over with! Simple, efficient, and straight to the point. Men!

[vava]

Correct me if I am wrong, but why in the world would you want to complicate a simple solution just multiply 6X4 and get it over with! Simple, efficient, and straight to the point. Men!

But then what happens to the 3 and the 1? They won't be able to join the foursome... so sad

 

Boghos - thanks, nice problem! I know, I'm a geek.

[Sasun]

Here is another probability problem, I have been trying to focus on it last 2 days but I seem to be unable to detach from other questions in HyeForum

 

This is a known problem, so if you know the solution please wait and give us a chance to solve.

 

There is a stick, you are randomly breaking the stick into 3 pieces. What is the probability that you can construct a triangle with the 3 pieces?

 

Note: there are geometrical solutions, if you can please provide a purely numeric solution.

[Sip]

What does randomly break into 3 pieces mean? Break into 2 random pieces and then pick one and break again? Or simultanously picking 2 non equal uniformly random values from 0 to L (where L is length) and break at those two points?

[Sasun]

What does randomly break into 3 pieces mean? Break into 2 random pieces and then pick one and break again? Or simultanously picking 2 non equal uniformly random values from 0 to L (where L is length) and break at those two points?

Don't know. I was given the problem without specification. I am assuming it should not matter but I was trying to solve for both breaks happening randomly anywhere (2nd of your case).

[Sasun]

I take it back, I think the method does matter. Lets try to solve for both cases.

[Sip]

Well, as a start, I guess we are trying to find the probability that given 3 numbers a,b,c the triangle inequality holds ... i.e. the sum of any two is greater than the third. So now the problem is modeling the relationship between a, b, and c with respect to L, the length of the stick.

 

If a and b are uniformly randomly selected from [0 to L], then c = L-max{a,b} ... I guess if this type of a specification is ok, then we can start thinking about writing the probabilities.

 

 

Edit: note the update in computing c.

Edited August 20, 2004 by Seapahn

[ED]

Here is another probability problem, I have been trying to focus on it last 2 days but I seem to be unable to detach from other questions in HyeForum

 

This is a known problem, so if you know the solution please wait and give us a chance to solve.

 

There is a stick, you are randomly breaking the stick into 3 pieces. What is the probability that you can construct a triangle with the 3 pieces?

 

Note: there are geometrical solutions, if you can please provide a purely numeric solution.

Sasun I think there should be more information, no? With only 3 pieces you are able to construct a triangle, what are you talking about 3 equal pieces? Or any lengths, 90, 45, 45 triangle? Be clearer please.

 

if you are breaking it equal or any lenght the probobilty is 100%

[Sasun]

Sasun I think there should be more information, no? With only 3 pieces you are able to construct a triangle, what are you talking about 3 equal pieces? Or any lengths, 90, 45, 45 triangle? Be clearer please.

 

if you are breaking it equal or any lenght the probobilty is 100%

Any triangle Edo. But it is not 100% because not all cases of breaking will form a triangle. See Sips post, which is also equivalent to say that each of 3 pieces should be shorter than half the size of the stick.

[Sasun]

If a and b are uniformly randomly selected from [0 to L], then c = L-max{a,b} ... I guess if this type of a specification is ok, then we can start thinking about writing the probabilities.

I am not sure, I was thinking c = L - a - b where c < L/2, b < L/2 and a < L/2.

 

The tricky thing is that the probabilities are conditional. But the first case that you mentioned makes it easier.

[ED]

ok you end up with a pieces thst is .875", .5", .375" how are those not making a triangle?

[DominO123]

ok you end up with a pieces thst is .875", .5", .375" how are those not making a triangle?

I have no time to meditate about the problem, but will just answer Ed.

 

Ed, if two pieces were by their sommes equal to the 3rd one, the angle between them would be 180, and the other angles being 0 and 0. If the two pieces are by their sommes even lower than the 3rd one, even when they are at 180 degree, they won't cover the distance of the 3rd one. So in conclusion, 2 of the pieces should have a somme of > the third one.

[Sip]

Thinking about it some more, I guess we want to find the probability that given the three pieces a, b, and c, we have a+b

 

We know a+b+c=L.

 

As a necessary but not sufficient condition, we have a+b

 

To make it a sufficient condition, we have that c>L/2.

 

So:

 

P(a+bL/2)

 

In other words, what are the chances that one of the broken pieces will be longer than L/2.

 

Case 1:

-------

We break the stick once, then break one of the pieces again. In this case, P(no triangle) = 1/2

 

Case 2:

-------

We pick two quantities x and y uniformly from [0..L] and break there. We will have c>L/2 if both [ xL/2 AND y>L/2]. In other words, both x and y are shorter than 1/2 L or longer than 1/2 L. It is easy to see that again the probability is 1/2.

 

Sounds like the probability of having or not having a triangle is 1/2.

 

Ok, now you smart guys point out all my mistakes

 

---

Edit: I think I did my math wrong. Unfortunately, not time to check now

Edited August 20, 2004 by Seapahn

[nairi]

Just a little suggestion: why don't we merge this thread with the math thread?

[Sip]

I stand corrected ... at least for case 2, I think the probabilites are more like 1/4 and 3/4. I am checking myself as of now.

[ED]

Sip i just drew in my comuper a tringle exacly with those numbers, I still maybe missing something?

[Sip]

Ed, I think you are misunderstanding the problem. Take a stick, break it into 3 pieces. If one of the pieces is longer than half of the original stick, there's no way you can form a triangle with the three pieces.

 

So the problem boils down to calculating the probability of having one of the broken pieces be longer than 1/2 the length of the original stick.

 

My problem is understanding how the breaking is done or if it matters at all.

[Sasun]

Just a little suggestion: why don't we merge this thread with the math thread?

Didn't the math thread merge with a romance thread long time ago ?

I also thought that there was a math thread but could not find it after a little search.

[Sasun]

I will be checking after I get something to eat, or this weekend. Enjoy in the meantime

[Sasun]

I stand corrected ... at least for case 2, I think the probabilites are more like 1/4 and 3/4.  I am checking myself as of now.

If you mean prob. triangle is 1/4 and prob. no triangle is 3/4 then I came to the same conclusion Furthermore, I am again thinking that the breaking method does not matter because the 2 breaks are independent regardless of the timing. In other words if you have to toss 2 coins it doesn't matter if you toss them at the same time or consecutively (tossing coins is my weekness ). The combined odds are the same in either case.

 

Now about the stick: let's fix the midpoint of the stick and identify all possibilities of breaking points for break 1 and break 2 (x and y) relative to the midpoint. There are four possibilities (I will be using Sip's notation above to avoid confusion):

 

Event 1. x and y are to the right of the midpoint. This is a no triangle case. For each x and y to be to the right the probability is 1/2. Since they are independent events then the probability of event 1 is 1/2*1/2=1/4

 

Event 2. x and y are to the left of the midpoint. Same as above, no triangle case with prob=1/4.

 

Event 3. x to the right and y to the left of the midpoint. Possibility of triangle but needs to satisfy certain constraints for sure triangle. Prob=1/4

 

Event 4. y to the right and x to the left of the midpoint. As in Event 3 possibility of triangle but needs to satisfy certain constraints. Prob=1/4

 

We can see that probabilities of Event 1 through event 4 add up, just to be sure

 

Now lets examine Event 3. Please draw a line so it can be seen clearly. In order for any 2 pieces of the 3 broken pieces the sum to be longer than the third piece, the condition below must satisfy.

 

From left to right we have piece a (length=y), piece b (length=x-y), piece c(length=L-x). It is trivial that any 2 adjasant pieces together are longer than the third piece. The only thing to determine is under what condition the pieces a and c are together longer than b. Let's see

 

y + (L-x) > x - y

 

which means x-y < L/2 .

 

On the other hand, event 3 implies that 0 is the lower bound of x-y. So, the probability of x - y being between 0 and L/2 is 1/2. We multiply this by the probability of event 3 and get 1/8.

 

The same analysis holds true for event 4, and we get additional 1/8 chance of a triangle. Therefore

 

1/8 + 1/8 = 1/4

 

will be the answer to the problem. Obviously, in 75% of the cases we will not get a triangle by randomly breaking a stick.

 

... man, its 3 am

Edited August 21, 2004 by Sasun

[Sip]

Sasun, I like your reasoning a lot. I have also verified it through simulation. For the case where the two break points are picked independently and uniformly from the range [0,L], the probability of the triangle inequality holding is 1/4.

[Sasun]

Simulation is cool Sip, I didn't think of it. It is a more reliable proof. So after all this question could be also a computer related puzzle

[Sip]

Here's my quick and dirty C code for anyone interested. The beauty of such a method is that it takes less than 10 minutes to set up and run while thinking through the math took at least an hour for me. In the code, I pick two uniform random values a and b from [0 to 1). Then I compute the segment lengths a,b,c. Then compute the triangle inequality and check to see if it holds. I have to do the inequality calculation 3 times depending on whether a,b,or c is the longest piece.

 

#include <iostream.h>
#include <stdlib.h>
#include <time.h>

int main() {
 srand(time(NULL));
 long not_tri=0;
 double a,b,c, tri;
 for (long i=0; i<1000000; i++) {
   a=(double)(rand()%RAND_MAX)/RAND_MAX;
   b=(double)(rand()%RAND_MAX)/RAND_MAX;
   
   if (a>b) { c=1-a; a=a-b; }
   else     { c=1-b; b=b-a; }
   
   tri=a-(b+c);
   if (b>a) tri=b-(a+c);
   if (c>b && c>a) tri=c-(a+b);
   
   if (tri>0) not_tri++;
 }
 cout << "Not triangles: " << (double)not_tri/1000000.0 << endl;
 return 0;
}

Edited August 21, 2004 by Seapahn

[Sasun]

Sip, looks like you did 1 million iterations. I am just curious how close you got to the numbers.

[Sip]

Here's the result from 10 runs:

 

Not triangles: 0.750357

Not triangles: 0.749882

Not triangles: 0.749753

Not triangles: 0.750702

Not triangles: 0.749753

Not triangles: 0.750702

Not triangles: 0.750337

Not triangles: 0.750095

Not triangles: 0.750819

Not triangles: 0.749169

 

median: 0.7502

mean: 0.7502

Variance: 2.7630e-007

Std Dev: 5.2565e-004

 

 

I'd say it's convincingly 3/4.

Edited August 21, 2004 by Seapahn